A solenoid 15 cm long has 300 turns of wire. A current of 5 A flows through it. What is the magnitude of the magnetic field inside the solenoid?
Correct answer: C. 4π x 10-3
- A. 75 x 10^7 T
- B. 60 x 10+3 T
- C. 4π x 10-3
- D. 750π x 10+3
Explanation
The magnetic field inside a solenoid is given by the formula:B=μ0nIwhere:B = Magnetic field (T)μ0 = Permeability of free space (4π×10−7n = Number of turns per unit length (N/L)I = Current (A)Given data:N=300N = 300N=300 turnsL=15L = 15L=15 cm = 0.15 mI=5I = 5I=5 AFirst, calculate n:n=NL=3000.15=2000 turns/mNow, calculate BBB:B=(4π×10−7)×(2000)×(5)Approximating π≈3.1416B≈4×3.1416×10−3
Last updated
About Magnetic Flux Density
Magnetic flux density measures the magnetic force acting on a current carrying conductor or moving charge in a magnetic field. Work includes the relations B equals F divided by IL and F equals qvB, the tesla as its SI unit, field direction and the distinction between flux density and magnetic flux.
Practise Electromagnetism
1,176 free Electromagnetism MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1 Gauss is equivalent to:
1 Weber / m² is equal to __________?
A 100m long conductor carrying current of 2A is at right angle to B of 0.5 wb_m² . The force experienced by the conductor is:
A 2m wire carrying current 5A is at right angle to uniform magnetic field of 0.2 Web/m². The force on wire will be:
A current carrying coil is placed in a uniform magnetic field.The torque produced in this coil will be doubled when the